Consider the previous problem , let the outer shell have the charge ` - 4 q`. As in the above problem , the inner shell has the charge ` + 2q`. Calculate the electric field in terms of `q` and the distance `r` from the common center of the two shells for
`c lt r lt d`

Options

A. zero
B. `(1)/( 4 pi epsilon_(0)) ( 2q)/( r^(2))`
C. `-(1)/( 4 pi epsilon_(0)) ( 2q)/( r^(2))`
D. `(1)/( 4 pi epsilon_(0)) ( 6 q)/( r^(2))`

Explanation

To solve the problem, we need to calculate the electric field in the region between the inner shell and the outer shell, specifically for the condition \( c < r < d \).

### Step-by-Step Solution:

1. **Identify the Charges**:
- The inner shell has a charge of \( +2q \).
- The outer shell has a charge of \( -4q \).

2. **Understand the Charge Distribution**:
- The inner shell will induce a charge of \( -2q \) on the inner surface of the outer shell due to electrostatic induction.
- Since the outer shell has a total charge of \( -4q \), the outer surface of the outer shell will have a charge of \( -4q + 2q = -2q \).

3. **Choose a Gaussian Surface**:
- We will choose a spherical Gaussian surface of radius \( r \) such that \( c < r < d \). This surface is located between the inner and outer shells.

4. **Apply Gauss's Law**:
- According to Gauss's Law, the electric flux through the Gaussian surface is given by:
\[ \Phi_E = \oint \vec{E} \cdot d\vec{S} = \frac{Q_{\text{enc}}}{\epsilon_0} \]
- Here, \( Q_{\text{enc}} \) is the charge enclosed by the Gaussian surface.

5. **Calculate the Enclosed Charge**:
- In the region \( c < r < d \), the only charge enclosed by the Gaussian surface is the charge of the inner shell, which is \( +2q \).
- The charge on the inner surface of the outer shell does not contribute to the enclosed charge since it is outside the Gaussian surface.

6. **Calculate the Electric Field**:
- Substituting \( Q_{\text{enc}} = +2q \) into Gauss's Law:
\[ \oint \vec{E} \cdot d\vec{S} = E \cdot 4\pi r^2 = \frac{2q}{\epsilon_0} \]
- Solving for \( E \):
\[ E = \frac{2q}{4\pi \epsilon_0 r^2} \]
- Therefore, the electric field in the region \( c < r < d \) is:
\[ E = \frac{q}{2\pi \epsilon_0 r^2} \]

### Final Answer:
The electric field in the region \( c < r < d \) is:
\[ E = \frac{q}{2\pi \epsilon_0 r^2} \]